1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Function Definition Without a Single Word

  1. Mar 18, 2012 #1
    Is this correct?

    A function ψ:A --> B is the set:

    ψ = { (x, y) | [itex]\forall[/itex]x[itex]\in[/itex]A[itex]\exists[/itex]y[itex]\in[/itex]B[itex]\ni[/itex](((x, y) [itex]\in[/itex] ψ) [itex]\wedge ((x, z)\in[/itex] ψ) [itex]\Rightarrow[/itex] y = z)}

    Thanks.
     
  2. jcsd
  3. Mar 18, 2012 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    No. Your definition is circular. Your definition of [itex]\psi[/itex] already uses [itex]\psi[/itex]. This is not allowed.
     
  4. Mar 18, 2012 #3
    I see what you saying but this is not a circular although it looks like so. The use of the same set [itex]\psi[/itex] in its definition is not for recursion.
     
  5. Mar 18, 2012 #4
    You're already in trouble here because in set-builder notation you have to say what set your elements are in. For example you need to write

    ψ = { (x, y)[itex]\in[/itex] something | ... }

    and that something is a key part of the definition that you're missing.

    (edit) This notational difficulty is actually related to the circularity people are talking about. What is (x,y), exactly? Where does it live? You are saying it lives in ψ but we don't know what that is because you're trying to define it.
     
    Last edited: Mar 18, 2012
  6. Mar 18, 2012 #5
    ψ = { (x, y) | ∀x∈A∃y∈B∋(((x, y) ∈ ψ) ∧((x,z)∈ ψ) ⇒ y = z)}

    psi is the set of (x,y) such that for all x in A there exists a y in B with (x,y) in psi and if (x,z) is in psi then y = z

    You've defined psi as the objects that are in psi
     
  7. Mar 18, 2012 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Then what is it for? "Recursion" in a definition is perfectly legitimate but, as you say, this is not recursion. In order to use that definition, you have to know which elements are in [itex]\psi[/itex] and which are not. Which makes no sense in a definition of [itex]\psi[/itex].
     
  8. Mar 18, 2012 #7
    This is a way of saying that if there are two pairs in this set (psi) with first coordinates equal then the pairs have to be equal. So what it serves in the definition: if such two pairs happen to exist then...happen to exist where? in this set.

    Another example when we define a set of even integers: E = { x | x = 2n where n belongs to Z} notice that E is defined in terms of another set Z of which E is a subset, where it's valid to write E is a subset of Z.
     
  9. Mar 18, 2012 #8
    You're doubling down on your notation error. It's this error that is causing you to not understand that you haven't told us where (x,y) lives or even what it is. You're missing a critical piece of information that's essential to the definition of a function.
     
  10. Mar 18, 2012 #9
    This isn't quite the same, since Z is already defined before providing the definition of E.
    In your example, you are in the process of *defining* the set psi. You can think of it as the set psi not existing until the entire formula has been correctly parsed. When you try parsing the part that says ( x, y) in psi , you will not know how to interpret this, as psi does not (..yet ) exist.

    The only way your interpretation of the formula would make sense, is if psi had already been previously defined, and the set construction formula is offered as a formula ranging over the set representing the function ( then the formula will be true by definition, assuming that psi had already been properly defined ).
     
  11. Mar 18, 2012 #10
    Understood. So corrected definition? Or any reference to what is and is not allowed in set definition?
     
  12. Mar 18, 2012 #11
    Ok then I would correct it to:

    F = { (x1, y), (x2, z) | (x1, y) and (x2 z) belong to AxB and (x1 = x2 ==> y = z) }
     
  13. Mar 18, 2012 #12
    It's good that you mentioned the Cartesian product. But you still have to account for the fact that there are a lot of different function from A to B.

    Also your notation's gotten worse. Now you are saying that a function is a pair of ordered pairs?

    It might help to work out some simple examples between finite sets. This problem's trickier than it looks.
     
    Last edited: Mar 18, 2012
  14. Mar 18, 2012 #13
    Incorrect, since I'm giving a general definition of a function/mapping.

    LoL. Then does AxB = { (a, b) | ... } make it consists of only one pair? :D
    Adding two pairs or 100 pairs does make no difference since the set will not allow duplicate elements.

    Then show me a definition that works. And working out some examples in mathematical logic may help. The problem is simpler than you think. :)
     
  15. Mar 18, 2012 #14

    jgens

    User Avatar
    Gold Member

    You used the notation {(x1,y)(x2,z) | blah blah blah} as your definition of a function. This means that elements of the function are pairs of ordered pairs. This is most certainly not the case.

    Evidently not. Otherwise you would have produced a correct definition by now.
     
  16. Mar 18, 2012 #15

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Are you here to learn something from us or to betittle and argue with us??
     
  17. Mar 18, 2012 #16
    The show me your correct definition please.

    To argue and learn. But you cannot tell me this is wrong without justifying your answer or at least give a reference to it. That's why I'm "arguing" or at least attempting to give a different answer even though not sure if it's correct.
     
  18. Mar 18, 2012 #17

    jgens

    User Avatar
    Gold Member

    That defeats the learning process. I am very firmly of the belief that in order to learn something you need to struggle with it and get it wrong many times before you can really understand it.

    Whenever someone has told you that your definition is wrong, they have pointed out at least one flaw with it. That is justification for their claim. Justification does not always need to come in the form of handing you the answer on a silver platter.
     
  19. Mar 18, 2012 #18
    I gave a counter example to the arguement:

    Where I said the AxB definition contains one pair { (a, b) | ... } and this does not mean it's only one element set. However no one proved me wrong.

    And I'm not expecting a solution to my homework, but a good but lazy teacher can give a reference to his smart-a** students if they are not convinced. :)
     
  20. Mar 18, 2012 #19

    jgens

    User Avatar
    Gold Member

    But that is not what you said. In your last attempt at defining a function, you wrote {(x1,y)(x2,z) | blah blah blah}. SteveL27 pointed out that this means that elements of your function are pairs of ordered pairs, which is not true. Sure he could have phrased his objection a little more clearly, but I tried to clear this confusion up with my first reply. If you did not read that part of my reply, surely no one helping you can be faulted.

    What kind of reference do you want? Whenever you have posted a definition, someone has pointed out some flaw with your definition and given hints on how you could fix it. Whenever you have raised objections to something, someone has suggested why your objections are a little silly. I am not sure what else you want.
     
  21. Mar 18, 2012 #20
    I'm here to ask, argue, suggest, try, without insulting other's intelligence and knowledge.
    And unless you give a clear argument supported by a mathematical law, I will not assume your are the Euler of the century and believe you blindly. All your arguments, and sorry, are just "NO" negative answers with one common flow, create confusion!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Function Definition Without a Single Word
  1. Definition of a word (Replies: 4)

Loading...